6) A random sam each hle of 30 households was selected from a particular neighbo

Question

6) A random sam each hle of 30 households was selected from a particular neighbon The number of ca 6) household is shown below. Estimate the mean namber of households in this neighborhood. Give-he 95% confiden interval. connumber of cars per household for the populatio 2 0123210 14 1320 1123 1 2 1041 022102 0)092.1)D)1.7) 7) The football coach randomly selected ten players and timed how long each player certain drill. The times (in minutes) were how long each player took to perform a 7__ 6.9 8.2 7.0 11.6 7.0 6.2 10.6 10.7 89 13.8 Determine a 95% confidence interval for the mean time for all players. A) (10.88,7.30) B) (7.40, 10.88) C) (7.30, 10.58) D) (10.78,7.40) E) (7.40, 10.78) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion. 8) Of 92 adults selected randomly from one town, 60 have health insuran e Ci snact a90% confidence interval for the percentage of all adults in the town who have health insurance. 8) A) (52.4%, 78.0%) B) (57.0%, 73.4%) C) (53.6%, 76.8%) D) (55.5%, 74.9%) E) (57.7%, 72.7%) 9) 9) When 252 college students are randomly selected and surveyed, it is found that115 own a car Construct a 99% confidence interval for the percentage of all college students who own-ar A) (40.5%, 50.8%) B) (39.5%, 51.8%) C) (38.3%, 52.9%) D) (37.6%, 53.7%) E) (25.3%, 66.0%) . i) Construct a 90% confidence interval for the percentage of all employees of the company who carpool. 10) Of 277 employees selected randomly from one company, 15.16% of them commute by carpooling A) (9.60%, 20.7%) B) (11.6%, 18.7%) C) (10.1%, 18.7%) D) (10.1 %, 20.2%) E) (10.9%, 19.4%)

Explanation / Answer

6) Correct answer: Option (A)
Margin of error, E = 0.4356094

95% Confident the population mean is within the range:
1.097724 < mean <1.968943

7)Correct answer: Option (C)
since
Margin of error, E = 1.785511

95% Confident the population mean is within the range:
7.304489 < mean <10.87551

8)Correct answer: Option (B)
since
Margin of error, E = 0.0816762

90% Confidence Interval (using normal approx):
0.5704977 < p < 0.7338501

9)Correct answer: Option (D)
Margin of error, E = 0.0808212
99% Confidence Interval (using normal approx):
0.3755281 < p < 0.5371704

10)Correct answer: Option (B)
Margin of error, E = 0.0354459

90% Confidence Interval (using normal approx):
0.1161787 < p < 0.1870704

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